Recall our usual cut-off function \(\varphi \in C^{\infty }_0(\mathbb {R}^2)\)  such that \(0\le \varphi \le 1\)  with \(\varphi \equiv 1\)  in \(B(1/2)\)  and \(\varphi \equiv 0\)  in \(B\setminus B(3/4)\) . Now define, for each \(n\in \mathbb {N}\) , the rescaled cut-off function \(\varphi _n(x) := \varphi (x/n)\) , and
\(\hat{\psi }^n_1 := (\psi _1 - [\psi _1]_{,n}) \varphi _n , \qquad \hat{\psi }^n_2 := (\psi _2 - [\psi _2]_{,n}) \varphi _n.\) 
We have that
\(v^n := (-\Delta )^{-1}(\nabla ^\perp \hat{\psi }^n_1 \cdot \nabla \hat{\psi }^n_2)\) 
is well-defined and
\(-\Delta v^n = \nabla ^\perp \hat{\psi }^n_1 \cdot \nabla \hat{\psi }^n_2 \qquad \text{in } \mathbb {R}^2.\) 
Our aim is now to obtain estimates on \(v^n\)  that are independent of \(n\) . We begin by observing that
\(\begin{aligned}\Vert \nabla \hat{\psi }^n_1 \Vert _{L_1(\mathbb {R}^2)} &\le \Vert \nabla \psi _1 \Vert _{L_1(\mathbb {R}^2)} + \frac{c(\varphi )}{n} \int _{B(n)} |\psi _1 - [\psi _1]_{,n}| \, \mathrm {d}x \\&\le c(\varphi ) \Vert \nabla \psi _1 \Vert _{L_1(\mathbb {R}^2)},\end{aligned}\) 
where we used the triangle inequality and the boundedness of \(\varphi \)  and its derivative to obtain the first inequality, and the Poincaré–Wirtinger inequality on a ball to obtain the second inequality. Similarly, we have that
\(\Vert \nabla \hat{\psi }^n_2 \Vert _{L_\infty (\mathbb {R}^2)} \le \Vert \nabla \psi _2 \Vert _{L_\infty (\mathbb {R}^2)} + \frac{c(\varphi )}{n}\sup _{x \in B(n)} \bigg | \psi _2(x) - \mathchoice{{\displaystyle {\textstyle -}{\int } }\hbox{$\textstyle -$ }}{\hspace{0.0pt}}{-}{.}6\) 



B(n) 2(y)   dy |,

where we used once again the triangle inequality and the boundedness of \(\varphi \)  and its derivative. Then, due to the bound on the gradient of \(\psi _2\) , we have that
\(\frac{1}{n}\sup _{x \in B(n)} \bigg | \psi _2(x) - \frac{1}{|B(n)|}\int _{B(n)} \psi _2(y) \, \mathrm {d}y \bigg | &\le \frac{1}{n}\sup _{x\in B(n)}\mathchoice{{\displaystyle {\textstyle -}{\int } }\hbox{$\textstyle -$ }}{\hspace{0.0pt}}{-}{.}6\) 



B(n)|2 (x) - 2(y)| dy
c2L(R2),       n N,

whence,
\(\Vert \nabla \hat{\psi }^n_2 \Vert _{L_\infty (\mathbb {R}^2)} \le c(\varphi ) \Vert \nabla \psi _2 \Vert _{L_\infty (\mathbb {R}^2)}, \qquad \forall n \in \mathbb {N}.\) 
Hence, the sequence \(\lbrace \nabla \hat{\psi }^n_1\rbrace _{n\in \mathbb {N}}\)  is uniformly bounded in \(L_1(\mathbb {R}^2)\) , and \(\lbrace \nabla \hat{\psi }^n_2\rbrace _{n\in \mathbb {N}}\)  is uniformly bounded in \(L_\infty (\mathbb {R}^2)\) .
In what follows, we directly estimate \(v^n\)  using its singular integral representation; these calculations are inspired from similar estimates due to Wente in [1]}. By rewriting the right-hand side of (REF ) in polar coordinates \((s,\theta )\) , we find that
\(\nabla ^\perp \hat{\psi }^n_1 \cdot \nabla \hat{\psi }^n_2 = \frac{1}{s}\partial _s (\hat{\psi }^n_1 \partial _\theta \hat{\psi }^n_2) - \frac{1}{s}\partial _\theta (\hat{\psi }^n_1 \partial _s \hat{\psi }^n_2),\) 
and hence, using the periodicity with respect to the angular variable,
we get that the expression for \(v^n(0)\)  reads as
\(v^n(0) = - \frac{1}{2\pi } \int _0^{2\pi } \bigg (\int _{0}^\infty \log s~ \partial _s (\hat{\psi }^n_1 \partial _\theta \hat{\psi }^n_2) \, \mathrm {d}s \bigg ) \, \mathrm {d}\theta ,\) 
where we also applied the Fubini-Tonelli theorem to change the order of integration. In turn, after an integration by parts, this becomes
\(v^n(0) = \frac{1}{2\pi } \int _0^{2\pi } \bigg (\int _{0}^\infty \hat{\psi }^n_1 \frac{1}{s} \partial _\theta \hat{\psi }^n_2 \, \mathrm {d}s \bigg ) \, \mathrm {d}\theta ;\) 
note that the boundary terms have vanished since
\(|\log {s}~ \hat{\psi }^n_1 \partial _\theta \hat{\psi }^n_2| \le \Vert \hat{\psi }^n_1\Vert _{L_{\infty }(\mathbb {R}^2)} \Vert \nabla \hat{\psi }^n_2\Vert _{L_{\infty }(\mathbb {R}^2)} |s \log s|\rightarrow 0\qquad \mbox{as }s\rightarrow 0^+,\) 
and because \(\hat{\psi }^n_1\)  and \(\hat{\psi }^n_2\)  are compactly supported. Using once again the Fubini-Tonelli theorem and the periodicity with respect to the angular variable in the \(\partial _\theta \hat{\psi }^n_2\)  portion of the integrand, the previous expression may be rewritten as
\(v^n(0) = \frac{1}{2\pi } \int _{0}^\infty \frac{1}{s} \bigg [\int _0^{2\pi } \big ( \hat{\psi }^n_1 - \frac{1}{2\pi }\int _0^{2\pi } \hat{\psi }^n_1(s,\alpha ) \, \mathrm {d}\alpha \big )\partial _\theta \hat{\psi }^n_2 \, \mathrm {d}\theta \bigg ] \, \mathrm {d}s.\) 
In turn,
\(\begin{aligned}|v^n(0)| &\le \frac{1}{2\pi } \Vert \nabla \hat{\psi }^n_2 \Vert _{L_\infty (\mathbb {R}^2)} \int _{0}^\infty \int _0^{2\pi } \bigg | \hat{\psi }^n_1 - \frac{1}{2\pi }\int _0^{2\pi } \hat{\psi }^n_1(s,\alpha ) \, \mathrm {d}\alpha \bigg | \, \mathrm {d}s \, \mathrm {d}\theta \\&\le c \Vert \nabla \hat{\psi }^n_2 \Vert _{L_\infty (\mathbb {R}^2)} \int _{0}^\infty \int _0^{2\pi } | \partial _\theta \hat{\psi }^n_1| \, \mathrm {d}s \, \mathrm {d}\theta \\&\le c \Vert \nabla \hat{\psi }^n_2 \Vert _{L_\infty (\mathbb {R}^2)} \Vert \nabla \hat{\psi }^n_1 \Vert _{L_1(\mathbb {R}^2)},\end{aligned}\) 
where we used the Poincaré-Wirtinger inequality to obtain the second inequality. By translation and using the uniform estimates (REF ) and (REF ), we obtain
\(\Vert v^n \Vert _{L_\infty (\mathbb {R}^2)} \le c \Vert \nabla \psi _2 \Vert _{L_\infty (\mathbb {R}^2)} \Vert \nabla \psi _1 \Vert _{L_1(\mathbb {R}^2)},\) 
for some positive constant \(c\) , independent of \(n\) . Additionally, by testing (REF ) with \(v^n \varphi _R\) , where \(\varphi _R\)  is our usual rescaled cut-off function, we obtain that
\(\int _{\mathbb {R}^2} |\nabla v^n|^2 \varphi _R \, \mathrm {d}x = \frac{1}{2} \int _{\mathbb {R}} |v^n|^2 \Delta \varphi _R \, \mathrm {d}x + \int _{\mathbb {R}^2} (\nabla ^\perp \hat{\psi }^n_1 \cdot \nabla \hat{\psi }^n_2) v^n \varphi _R \, \mathrm {d}x.\) 
We thereby deduce from the uniform estimates (REF )-(REF ) that there exists a positive constant \(c\)  independent of \(n\)  such that
\(\Vert \nabla v^n \Vert _{L_2(\mathbb {R}^2)} \le c \Vert \nabla \psi _2 \Vert _{L_\infty (\mathbb {R}^2)} \Vert \nabla \psi _1 \Vert _{L_1(\mathbb {R}^2)}.\) 
By the Banach-Alaoglu theorem, the uniform bounds (REF ) and (REF ) imply the existence of a subsequence (which we still label as \(\lbrace v^n\rbrace _{n\in \mathbb {N}}\) ) and a function \(v\)  such that
\(v^n \overset{*}{\rightharpoonup } v \quad \text{in } L_\infty (\mathbb {R}^2), \qquad \nabla v^n \rightharpoonup \nabla v \quad \text{in } L_2(\mathbb {R}^2).\) 
Let us return now to the equation (REF ). We introduce the following test function \(\zeta \in C^\infty _0(\mathbb {R}^2)\) ; there exists \(R>0\)  such that \(\operatornamewithlimits{supp}\zeta \subset B(R)\) . We find that
\(\int _{\mathbb {R}^2} \nabla v^n \cdot \nabla \zeta \, \mathrm {d}x = \int _{\mathbb {R}^2} (\nabla ^\perp \psi _1 \cdot \nabla \psi _2) \zeta \, \mathrm {d}x\) 
for all \(n \ge 2 R + 1\) . We note that arguing in this way avoids having to directly consider the product of two weakly convergent sequences; on the right-hand side of the above.
Using (REF ) to pass to the limit in the previous equation, we find that
\(\int _{\mathbb {R}^2} \nabla v \cdot \nabla \zeta \, \mathrm {d}x = \int _{\mathbb {R}^2} (\nabla ^\perp \psi _1 \cdot \nabla \psi _2) \zeta \, \mathrm {d}x \qquad \forall \zeta \in C^\infty _0(\mathbb {R}^2),\) 
and hence \(v\)  solves (REF ) in the sense of distributions. Estimate (REF ) follows from the convergences (REF ), estimates (REF )-(REF ) and the weak (resp. weak-*) lower semicontinuity of the norms.
